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Calculus II - Cheatsheet

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Section 1

Calculus II - Cheatsheet

STUDY GUIDE

🎓 Calculus II - Study Guide

📋 Course Structure

code
📚 Calculus II ├── 📖 Chapter 1: Integration by Parts │ ├── 🔹 The Integration by Parts Formula │ ├── 🔹 Integration by Parts: Definite Integrals │ └── 🔹 Reduction Formulas └── 📖 Chapter 2: Improper Integrals │ ├── 🔹 Improper Integrals of Type 1: Infinite Intervals │ ├── 🔹 Improper Integrals of Type 2: Discontinuous Integrands │ └── 🔹 Comparison Test for Integrals
Section 2

📖 Chapter 1: Integration by Parts

What this chapter covers: This chapter introduces the integration by parts technique, a method used to integrate products of functions. It covers the theoretical foundation, practical application, and definite integrals. Strategies for choosing appropriate parts to integrate are provided. Reduction formulas are also introduced as a specific application of integration by parts, useful for simplifying integrals iteratively.

🔑 Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to UseQuick Check
Integration by Parts∫f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dxIntegrating product of functionsDifferentiate result to check
Definite Integration by Parts∫ₐᵇ f(x)g'(x) dx = [f(x)g(x)]ₐᵇ - ∫ₐᵇ f'(x)g(x) dxDefinite integrals of productsEvaluate both sides independently
LIATELogarithmic, Inverse trig, Algebraic, Trig, ExponentialChoosing 'u' in integration by partsPrioritize left to right

🛠️ Problem Types

Type A: Indefinite Integral with Product of Functions
Setup: "When you see a product of functions like xsin(x) or x²e^x"
Method: Identify 'u' and 'dv' using LIATE. Apply the integration by parts formula.
Example: ∫x cos(x) dx = x sin(x) - ∫sin(x) dx = x sin(x) + cos(x) + C

Type B: Definite Integral with Product of Functions
Setup: "If given a definite integral of a product, like ∫₀^(π/2) x cos(x) dx"
Method: Apply integration by parts, then evaluate the result at the limits of integration.
Example: ∫₀^(π/2) x cos(x) dx = [x sin(x)]₀^(π/2) - ∫₀^(π/2) sin(x) dx = (π/2) - 1

🧮 Solved Example

Problem: Evaluate ∫x²e^x dx

Given: Integral of a product of algebraic and exponential functions.

"
Solution: 1. Let u = x², dv = e^x dx
  1. Then du = 2x dx, v = e^x
  2. ∫x²e^x dx = x²e^x - ∫2xe^x dx
  3. Apply integration by parts again: u = 2x, dv = e^x dx
  4. du = 2 dx, v = e^x
  5. ∫2xe^x dx = 2xe^x - ∫2e^x dx = 2xe^x - 2e^x + C
  6. Therefore, ∫x²e^x dx = x²e^x - 2xe^x + 2e^x + C
"
Answer: x²e^x - 2xe^x + 2e^x + C

⚠️ Common Mistakes

❌ Mistake 1: Incorrectly choosing 'u' and 'dv'
✅ How to avoid: Use LIATE to prioritize functions for 'u'.

❌ Mistake 2: Forgetting the constant of integration
✅ How to avoid: Always add "+ C" to indefinite integrals.

🦁 Erik's Tip

When applying integration by parts multiple times, keep track of your 'u' and 'dv' selections to avoid circular reasoning.

📖 Chapter 2: Improper Integrals

What this chapter covers: This chapter covers improper integrals, which involve infinite limits of integration or discontinuities within the integration interval. It defines two types of improper integrals and provides methods for evaluating them. Convergence and divergence of improper integrals are discussed, along with comparison tests to determine convergence without explicitly evaluating the integral.

🔑 Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to UseQuick Check
Improper Integral (Type 1)∫ₐ^∞ f(x) dx = lim(b→∞) ∫ₐᵇ f(x) dxInfinite limit of integrationCheck if limit exists
Improper Integral (Type 2)∫ₐᵇ f(x) dx = lim(t→b⁻) ∫ₐᵗ f(x) dx (f(x) discontinuous at b)Discontinuity within intervalCheck if limit exists
Comparison TestIf 0 ≤ g(x) ≤ f(x) and ∫ₐ^∞ f(x) dx converges, then ∫ₐ^∞ g(x) dx convergesDetermining convergence without direct integrationEnsure 0 ≤ g(x) ≤ f(x)

🛠️ Problem Types

Type A: Improper Integral with Infinite Limit
Setup: "When you see an integral with ∞ as a limit, like ∫₁^∞ 1/x² dx"
Method: Replace ∞ with 'b', take the limit as b→∞ of the definite integral.
Example: ∫₁^∞ 1/x² dx = lim(b→∞) ∫₁ᵇ 1/x² dx = lim(b→∞) [-1/x]₁ᵇ = 1

Type B: Improper Integral with Discontinuity
Setup: "If given an integral with a discontinuity within the interval, like ∫₀¹ 1/√x dx"
Method: Replace the limit of integration with 't' approaching the discontinuity, take the limit.
Example: ∫₀¹ 1/√x dx = lim(t→0⁺) ∫ₜ¹ 1/√x dx = lim(t→0⁺) [2√x]ₜ¹ = 2

🧮 Solved Example

Problem: Determine if ∫₁^∞ x/(1+x²) dx is convergent or divergent.

Given: Improper integral with an infinite limit.

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Solution: 1. ∫₁^∞ x/(1+x²) dx = lim(b→∞) ∫₁ᵇ x/(1+x²) dx
  1. Let u = 1+x², du = 2x dx
  2. (1/2)∫(1/u) du = (1/2)ln|u|
  3. (1/2)ln(1+x²) |₁ᵇ = (1/2)[ln(1+b²) - ln(2)]
  4. lim(b→∞) (1/2)[ln(1+b²) - ln(2)] = ∞
"
Answer: Divergent

⚠️ Common Mistakes

❌ Mistake 1: Ignoring the limit when evaluating improper integrals
✅ How to avoid: Always replace infinite limits or points of discontinuity with a variable and take the limit.

❌ Mistake 2: Incorrectly applying the comparison test
✅ How to avoid: Ensure the inequality 0 ≤ g(x) ≤ f(x) holds and that you know the convergence/divergence of ∫g(x) dx or ∫f(x) dx.

🦁 Erik's Tip

When using the comparison test, try to find a function that behaves similarly to the integrand for large values of x.

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